Educational Codeforces Round 86 (Rated for Div. 2)（A&&B）

Road To Zero CodeForces - 1342A
You are given two integers x and y. You can perform two types of operations:

Pay a dollars and increase or decrease any of these integers by 1. For example, if x=0 and y=7 there are four possible outcomes after this operation:
x=0, y=6;
x=0, y=8;
x=−1, y=7;
x=1, y=7.
Pay b dollars and increase or decrease both integers by 1. For example, if x=0 and y=7 there are two possible outcomes after this operation:
x=−1, y=6;
x=1, y=8.
Your goal is to make both given integers equal zero simultaneously, i.e. x=y=0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.

Calculate the minimum amount of dollars you have to spend on it.

Input
The first line contains one integer t (1≤t≤100) — the number of testcases.

The first line of each test case contains two integers x and y (0≤x,y≤109).

The second line of each test case contains two integers a and b (1≤a,b≤109).

Output
For each test case print one integer — the minimum amount of dollars you have to spend.

Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391+555+391=1337 dollars.

In the second test case both integers are equal to zero initially, so you dont’ have to spend money.
思路：如果2*a<=b的话，就用第一种方法；否则共同的那一部分用b，多于的那一部分用a。
代码如下：

#include
#define ll long long
using namespace std;

ll x,y,a,b;

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		cin>>x>>y>>a>>b;
		if(x==y&&x==0) cout<<0<<endl;
		else
		{
			if(a*2<=b) cout<<(x+y)*a<<endl;
			else cout<<min(x,y)*b+(max(x,y)-min(x,y))*a<<endl;
		}
	}
	return 0;
}

Binary Period CodeForces - 1342B
思路：在2*t范围内，我们可以构造出101010…这样的串的，所以k如果不是1的话，那就是2.先检测一下原来的串是否只由一种元素构成，如果不是那就构造出10101010…这样的。
代码如下：

#include
#define ll long long
using namespace std;

string s;

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		cin>>s;
		int flag=1;
		for(int i=1;i<s.length();i++)
		{
			if(s[i]!=s[i-1])
			{
				flag=0;
				break;
			}
		}
		if(flag) cout<<s<<endl;
		else 
		{
			for(int i=0;i<2*s.length();i++)
			{
				if(i%2==0) cout<<1;
				else cout<<0;
			}
			cout<<endl;
		}
	}
	return 0;
}

努力加油a啊，(^o)/~